A 3.50 L balloon contains 0.178 mol of gas. If 0.094 mol of gas is

2. The color of blue light has an energy (E) of 6.62 x 10-19 J. What is the

malfutka [58]

67.45 is the answer I think

8 0

2 years ago

Determine the number of charged particles in nucleus of calcium atom then deduce the number of electrons

Dafna1 [17]

Answer:

detail is given below.

Explanation:

The charged particles of nucleus are protons while neutrons are neutral having no charge.

We know that an atom consist of electrons, protons and neutrons. Neutrons and protons are present inside the nucleus while electrons are present out side the nucleus.

Electron has a negative charge and is written as e⁻. The mass of electron is 9.10938356×10⁻³¹ Kg . While mass of proton and neutron is 1.672623×10⁻²⁷Kg and 1.674929×10⁻²⁷ Kg respectively.

Symbol of proton= P⁺

Symbol of neutron= n⁰

The number of electron or number of protons are called atomic number while mass number of an atom is sum of protons and neutrons.

one proton contribute 1 amu to the total weight. There are 20 protons and 20 neutrons in Ca thus its atomic mass is 40 amu.

While the atomic number is 20.

3 0

3 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

3 years ago

Please help ASAP please

77julia77 [94]

Answer:

3

Explanation:

Number of Energy Levels: 3

First Energy Level: 2

Second Energy Level: 8

Third Energy Level: 7

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5 0

2 years ago

Read 2 more answers

Reserve one spot for dmndburgess thx

Damm [24]

Answer:

thanks for the points have a nice day :D

Explanation:

4 0

3 years ago